Example 4
Completion requirements
Example 4 |
Simplify \(\frac{2}{{\sqrt[4]{{4r}}}}, r > 0\).
This time, in order to remove the fourth root from the denominator, you must change the radicand so that it has an exponent of four. Therefore, multiply both numerator and denominator by \(\left( {\sqrt[4]{{4r}}} \right)^3 \).
\(\begin{align}
\frac{2}{{\sqrt[4]{{4r}}}} &= \frac{2}{{\sqrt[4]{{4r}}}}\cdot \frac{{\left( {\sqrt[4]{{4r}}} \right)^3 }}{{\left( {\sqrt[4]{{4r}}} \right)^3 }} \\
&= \frac{{2\left( {\sqrt[4]{{4r}}} \right)^3 }}{{\sqrt[4]{{\left( {4r} \right)^4 }}}} \\
&= \frac{{2\sqrt[4]{{64r^3 }}}}{{4r}} \\
&= \frac{{2\sqrt[4]{{16 \cdot 4r^3 }}}}{{4r}} \\
&= \frac{{2 \cdot 2\sqrt[4]{{4r^3 }}}}{{4r}} \\
&= \frac{{\sqrt[4]{{4r^3 }}}}{r} \\
\end{align}\)
\(\begin{align}
\frac{2}{{\sqrt[4]{{4r}}}} &= \frac{2}{{\sqrt[4]{{4r}}}}\cdot \frac{{\left( {\sqrt[4]{{4r}}} \right)^3 }}{{\left( {\sqrt[4]{{4r}}} \right)^3 }} \\
&= \frac{{2\left( {\sqrt[4]{{4r}}} \right)^3 }}{{\sqrt[4]{{\left( {4r} \right)^4 }}}} \\
&= \frac{{2\sqrt[4]{{64r^3 }}}}{{4r}} \\
&= \frac{{2\sqrt[4]{{16 \cdot 4r^3 }}}}{{4r}} \\
&= \frac{{2 \cdot 2\sqrt[4]{{4r^3 }}}}{{4r}} \\
&= \frac{{\sqrt[4]{{4r^3 }}}}{r} \\
\end{align}\)